Proof of Continuity of $x^n$ from Pugh's "Real Mathematical Analysis" Chapter 1

[Someone2841]

From Pugh's "Real Mathematical Analysis" Chapter 1

13. Given $y \in \mathbb{R}, n \in \mathbb{N}, \text{ and } \epsilon > 0$, show that for some $\delta > 0, |y-x| < \delta \text{ then } |y^n - x^n| < \epsilon$

In the case of n=1, $\delta = \epsilon$ satisfies the condition, i.e. $|y-x| < \delta = \epsilon \implies |y - x| < \epsilon$.

In the case of n=2, it needs to be shown that $|y-x| < \delta \implies |y-x||y+x| < \epsilon$

1. $|y+x| < 1$ implies that $|y-x||y+x| < |y-x| < \delta$. If $\delta = \epsilon$, then the condition is met.

2. $|y+x| \geq 1$ implies that $|y-x| \leq |y-x||y+x| < \delta \text{ } |y+x|$

Here's my question: Is it valid to set $\delta$ to $\frac{\epsilon}{|y+x|}$? In the case that $|y+x| \geq 1$, there is no risk of division by zero or assigning $\delta$ to a negative value, so $\delta$ would exist and be positive. If so, then the condition is met.​

[Note: This is only the first part of the problem; I do intend to solve the general case with induction.]


Thanks!

 

[micromass]

I'm afraid it is not valid. You want your $\delta$ to be independent of $x$. This means that you want to find a $\delta>0$ such that your implication holds for all $x$.
If you choose $\delta = \varepsilon / |x+y|$, then you will get a different value of $\delta$ for each $x$. Clearly this is not what you want.

 

[Someone2841]

I'm afraid it is not valid. You want your $\delta$ to be independent of $x$. This means that you want to find a $\delta>0$ such that your implication holds for all $x$.
If you choose $\delta = \varepsilon / |x+y|$, then you will get a different value of $\delta$ for each $x$. Clearly this is not what you want.

That makes sense. Does this work?

2. $|y+x|≥1$ implies that $|y−x|≤|y−x||y+x|<ε.$ Let $δ=ε$, then the condition is met?

If it does, I've been making it out to be more complicated than it is.

Thanks!

 

[Someone2841]

1. $|y+x| < 1$ implies that $|y-x||y+x| < |y-x| < \delta$. If $\delta = \epsilon$, then the condition is met.

Is this alright? I notice that there is the possibility that $\delta = \epsilon< |y-x|$ for certain x, y, and ε, which would cause $|y^2-x^2| < \epsilon$ but $|y-x| \geq \delta$; however, $|y-x| < \delta \implies |y^2-x^2| < \epsilon$ requires only that $|y^2-x^2| < \epsilon$ is true whenever $|y-x| < \delta$ and not the other way around, so it seems okay to me. Am I correct?

Example: $x=0.1, y=0.2, δ=ε=.04$

$.1 < .04 \implies .03<.04 \text{ is true.}$​

Thanks!

 

[Someone2841]

13. Given $y∈R,n∈N, \text{ and } ϵ>0$, show that for some $δ>0,|y−x|<δ \text{ then } |y^n−x^n|<ϵ$

For n>2, the following formula is helpful: $|y^n-x^n| = |y-x||y^{n-1} + xy^{n-2}+\cdots+x^{n-2}y+x^{n-1}|$

It seems to me that, provided my methods for n=2 were valid, that the same arguments could be used for the cases $|y^{n-1} + xy^{n-2}+\cdots+x^{n-2}y+x^{n-1}| < 1$ and $|y^{n-1} + xy^{n-2}+\cdots+x^{n-2}y+x^{n-1}| \geq 1$ to show that $\delta = \epsilon$ satisfies all conditions for continuity.

 

[deluks917]

I think it is easier to do this problem if you prove {lim}_{h \rightarrow 0} |(x+h)^n -x^n| = 0.